Description
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a textT, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
- One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
- One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
题意:在第二个字符串中查找第一个字符串出现的个数。
以第一个字符串建立next函数,然后在第二个字符串中查找。
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 int next[100000]; 5 char s1[100000],s2[2000000]; 6 void get_next() 7 { 8 int i=0,j=-1; 9 next[0]=-1; 10 while (s1[i]) 11 { 12 if (j==-1||s1[i]==s1[j]) 13 { 14 i++; 15 j++; 16 next[i]=j; 17 } 18 else j=next[j]; 19 } 20 } 21 int kmp() 22 { 23 int i=0,j=0,k=0,len; 24 len=strlen(s1); 25 while (s2[i]) 26 { 27 if (j==-1||s1[j]==s2[i]) 28 { 29 j++; 30 i++; 31 } 32 else j=next[j]; 33 if (j==len) 34 { 35 k++; 36 j=next[j]; 37 } 38 } 39 return k; 40 } 41 int main() 42 { 43 int n; 44 scanf("%d",&n); 45 getchar(); 46 while (n--) 47 { 48 gets(s1); 49 gets(s2); 50 memset(next,0,sizeof(next)); 51 get_next(); 52 int ans=kmp(); 53 printf("%d ",ans); 54 } 55 }